CH 39 #49 - 39.49 a)  ∆p(min = h 6.63 × 10 −34 J ⋅...

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Unformatted text preview: 39.49: a)  ∆p (min) = h 6.63 × 10 −34 J ⋅ s = = 2.1 × 10 −20 kg ⋅ m s −15 2π∆x 2π (5.0 × 10 m) b)  E = ( pc) 2 + (mc 2 ) 2 = [(2.1 × 10 −20 kg ⋅ m s)(3.0 × 10 8 m s)]2 + [(9.11 × 10 −31 kg)(3.0 × 10 8 m s) 2 ] 2 = 6.3 × 10 −12 J = 39.5MeV. K = E − mc 2 = 38.8 MeV   c) The coulomb potential energy is U = q1q2 (1.60 × 10 −19 C) 2 ⇒U = − = 4πε0V 4πε0 (5.0 × 10 −15 m) − 4.60 × 10 −14 J = −0.29 MeV Hence there is not enough energy to “hold” the electron in the nucleus. ...
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